If the system of equations 2x+3y-z=5 x+ y+3z=-4 3x-y+ z=7 has infinitely many solutions, then 13 is equal to :
- 1110
- 1120
- 1210
- 1220
Answer
(2) 1120
Solution
For infinitely many solutions the three planes must pass through one common line, so the first plane belongs to the family through the other two:
(2x + 3y - z - 5) = k1(x + y + 3z + 4) + k2(3x - y + z - 7)
Comparing coefficients: x: 2 = k1 + 3k2; constant: -5 = 4k1 - 7k2
These two give k2 = 1319, k1 = - 119
y: 3 = k1 - k2 57 = - - 13 = -70
z: -1 = 3k1 + k2 -19 = -3 + 13 = - 1613
13 = 13(-70) (- 1613 ) = 1120
Answer: (2)